tchrwill

Most popular questions and responses by tchrwill
  1. Algebra

    A movie theater manager wants to know how many adults and how many children pay admission to a particular movie. The theater charges $10.00 for adult tickets and $5.00 for child tickets. At a showing where 210 tickets were sold the theater collected

    posted on July 21, 2015
  2. Physics

    The equations of accelerated motion, derived elsewhere, also apply to falling (or rising) bodies with the exception that the term "a" for acceleration is replaced by the term "g” for the acceleration due to gravity). This results in ....Vf = Vo + gt (the

    posted on April 14, 2013
  3. Geometry

    The area of a spherical triangle bounded by arcs of three great circles is defined by A = Pi(r^2)E/180 where r = the spherical radius and E = the spherical excess, defined by E = (A+B+C)-180, A, B and C being the three spherical angles of the sphere.

    posted on March 6, 2013
  4. wireless communications

    The velocity required to maintain a circular orbit around the Earth may be computed from the following: Vc = sqrt(µ/r) where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of

    posted on December 2, 2012
  5. Physics

    Using the radius of the Earth and Moon as 3963 miles and 1080 miles, respectively, and the distance between them as 239,000 miles, the surface to surface distance becomes 233,960 miles. From Vf^2 = Vo^2 - 2gs, 0 = Vo^2 - 2(32.2)233,960 or Vo = 282,053 fps

    posted on November 16, 2012
  6. Physics

    The velocity required to maintain a circular orbit around the Earth may be computed from the following: Vc = sqrt(µ/r) where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of

    posted on November 11, 2012
  7. physics

    Contrary to popular belief, a spacecraft does not have to reach full escape velocity in order to reach the Moon. The full escape velocity from a 200 mile high circular orbit is 24,400 mph. Assuming the trip starts from a 200 mile high circular orbit, the

    posted on October 24, 2012
  8. Physics

    There is a clear distinction between a geosynchronous orbit and a geostationary orbit. The early recognition of a geostationary orbit was made by the Russian Konstantin Tsiolkovsky early this century. Others referred to the unique orbit in writings about

    posted on October 17, 2012
  9. math

    The formula that will clear up your problem is ......Vc = sqrt(µ/r) where Vc = the velocity required to keep a body in a circular orbit, in feet/sec., r = the orbital radius in feet and µ = the Earth's gravitational constant or 1.407974x10^16 ft^3/sec^2.

    posted on August 12, 2012
  10. Algebra

    2X + (X - 4) = 32 - X

    posted on July 29, 2012
  11. Physics

    Can you explain wheightlessness. Do you actually weigh nothing in space or just a percentage of your Earth weight? How much would a 100 lb person weigh on each of the other planets? >> An astronaut, circling the earth in the Space Shuttle, senses that his

    posted on July 9, 2012
  12. Physics

    There is a clear distinction between a geosynchronous orbit and a geostationary orbit. The early recognition of a geostationary orbit was made by the Russian Konstantin Tsiolkovsky early this century. Others referred to the unique orbit in writings about

    posted on July 1, 2012
  13. geometry

    * The altitude to the hypotenuse of a right triangle creates two similar triangles, each similar to the original right triangle and to each other. * The altitude to the hypotenuse of a right triangle is the geometric mean between the segments of the

    posted on May 25, 2012
  14. physics

    Since the velocity of the train is constant at 54kmph, the average velocity is 54kmph. Proof: D1 = 54(.5) = 27km. D1 = 54(.6666666) = 36 Vav = (27 + 36)/1.16666 = 54kmph.

    posted on May 25, 2012
  15. physics

    Time out = 6/2.5 = 2.4 hours Time back = 6/4 = 1.5 hours Average speed for the round trip = (6 + 6/(2.4 + 1.5) = 3.077 kmph

    posted on May 25, 2012
  16. inscribed angles

    Clearly, I did not understand you. But after reviewing the scenario, the answer is clearly d = sqrt(17000^2 + 4000^2)-4000 = 13,464

    posted on May 11, 2012
  17. Math

    The numbers 4; 6; 13; 27; 50; 84 do not form an arithmetic progression as the differences between succsessive terms are not constant. If you take the successive differences of the terms given, n.......1....2....3....4....5....6...

    posted on May 11, 2012
  18. inscribed angles

    If I understand you correctly the distance you seek is d = sqrt(17000^2 - 4000^2) - 4000

    posted on May 11, 2012
  19. Physics

    Given the three sides (the largest must be less than the sum of the other two) Find any angle from cosA = (b^2 + c^2 - a^2)/2bc. The remaining angles can be derived from the Law of Sines. Alternatively, find any angle, A for instance, using tan(A/2) = r/(s

    posted on May 6, 2012
  20. physics

    Moment of inertia M(i) = 2WR^2/5 about any axis.

    posted on April 23, 2012
  21. physics

    Thye following is more than you asked for but I hope you find it interesting. I think you would be surprised to know just how many people have no idea that the Earth moves at all, thinking that the Sun revolves around the Earth. The motion of our Earth

    posted on April 18, 2012
  22. Geometry

    3.14(11^2)/6 - 2(11)5.5/2 = 63.355 - 60.5 = 2.855 sqcm Therefore, D.

    posted on April 16, 2012
  23. Algebra

    Let A = the number of adult tickets C = the number of children tickets Then, A + C = 210 and 10A + 5C = 1355 Can you take it from here?

    posted on March 26, 2012
  24. trig

    Part of my previous reply was lost in the posting process. The orbital radius is 3960 + 2.1(104) = 4178.4 miles or 22,061,952 feet. The alleged time to complete one orbit is 10.5(24)3600 = 907,200 seconds making the derived orbital velocity Vc = 152.7fps.

    posted on March 19, 2012
  25. trig

    /r] sqrt(sqrt[(1.407974x10^16)/(3960+218.4)5280] = 25,262 feet per second. The orbital period is T = 2(3.14)sqrt[22,061,952^3/1.407974x10^16] = 5487 seconds or 91.45 minutes.

    posted on March 19, 2012
  26. algebra

    Continuous Compound Interest Formula where FV = Pe^(rt) P = principal amount (initial investment) r = annual interest rate (as a decimal) t = number of interest bearing years FV = amount after time t 110,682 = 45,000e^(30r) from which r = .03 or an annual

    posted on March 19, 2012
  27. physics

    As we normally think of it, the static value of gravity on, or above the surface of a spherical body is directly proportional to the mass of the body and inversely proportional to the square of the distance from the center of the body and is defined by the

    posted on March 14, 2012
  28. Physics

    An asteriod is found a distance from the sun equal to 32 units times the earth's distance. What will be the length of time in years required for this asteroid to make one revolution around the sun? Rast = 32 x 92,960,242 = 2,974,727,744 miles GMsun =

    posted on March 12, 2012
  29. physics

    The acceleration due to gravity derives from g = µ/R^2 where G = the acceleration due to gravity µ = the earths gravitational constant = GM G = the universal gravitational constant = 6.67259x10^-11 M = the mass of the earth = 5.97424x10^24 R = the mean

    posted on December 15, 2011
  30. physics

    When orbiting at 5.8km/s: From Vc = sqrt(µ/r) where Vc = the velocity of an orbiting body, µ = the gravitational constant of the earth and r the radius of the circular orbit,with µ = GM, G = the universal gravitational constant and M = the mass of the

    posted on December 11, 2011
  31. physics

    When orbiting at 6.3km/s: From Vc = sqrt(µ/r) where Vc = the velocity of an orbiting body, µ = the gravitational constant of the earth and r the radius of the circular orbit,with µ = GM, G = the universal gravitational constant and M = the mass of the

    posted on December 10, 2011
  32. physics

    g = µ/r^2 gravity on the surface of Mars µ = the gravitational constant of Mars = GM where G = the universal gravitational constant and M = the planet mass r = the radius of Mars = .53(6378)1000 = 3,380,340m G = 6.67259x10^-11 M = .11(5.97424x10^24) =

    posted on December 10, 2011
  33. algebra

    Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area? Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P -

    posted on December 10, 2011
  34. algebra

    What is the monthly deposit required to accumulate to a fund of $1,000,000 over a period of 40 years with deposits starting at the end of the first month and bearing an interest rate of 8% compounded monthly? S(n) = $1,000,000] i = .08/12 = .00666... n =

    posted on December 9, 2011
  35. Physics

    For Vex = velocity of the exhaust gases g = acceleration due to gravity w = the flow rate of the exhaust gases F = the thrust exerted on the rocket Vex/g = F/w or F = wVex/g

    posted on December 9, 2011
  36. math

    x(x + 1) = x + (x + 1) + 11 Your move.

    posted on December 9, 2011
  37. Math

    37 + 10A + B = 59 10A + B = 22 10A + 2 = 22 10A = 20 making A = 2 or 37 + 22 = 59

    posted on December 9, 2011
  38. math

    A convex polyhedron is defined as a solid with flat faces and straight edges so configured as to have every edge joining two vertices and being common to two faces. There are many convex polyhedra, only five of which are considered regular polyhedra.

    posted on December 7, 2011
  39. Physics

    The following might be of some help to you. 1/23/02 What is the final burnout velocity of the Space Shuttle? The final burnout velocity of the Space Shuttle, or any rocket for that matter, is a function of the altitude at burnout and the orbit that the

    posted on December 7, 2011
  40. science

    Assuming you are seeking the altitude where gravity = g = 6.5m/s^2: As we normally think of it, the static value of gravity on, or above the surface of a spherical body is directly proportional to the mass of the body and inversely proportional to the

    posted on December 7, 2011
  41. math

    Lets look at a typical problem with real numbers. If there are 15 people in a room and each person shakes hands with every other person in the room only once, how many handshakes will take place? There are two ways of looking at this type of problem. The

    posted on December 6, 2011
  42. Physics

    Vc = sqrt(µ/r) V(A)^2 = µ/r(A) V(B)^2 = µ/r(B) = µ/2r(A) V(B)^2/(2/V(A)^2 = [µ/2r(A)]/[µr(A)] .................= 2 V(B)/V(A) = sqrt2

    posted on December 5, 2011
  43. algebra

    x + y = 172 110x + 235y = 29,420. Can you take it from here?

    posted on December 3, 2011
  44. Physics

    From Vc = sqrt(µ/r) where Vc = the velocity of an orbiting body, µ = the gravitational constant of the earth and r the radius of the circular orbit,with µ = GM, G = the universal gravitational constant and M = the mass of the central body, the earth in

    posted on December 2, 2011
  45. Algebra

    Considering all rectangles with the same perimeter, the square encloses the greatest area. Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side

    posted on December 2, 2011
  46. physics

    The velocity required to maintain a circular orbit around the Earth may be computed from the following: Vc = sqrt(µ/r) where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of

    posted on November 29, 2011
  47. math

    The present value of an ordinary annuity is the sum of the present values of the future periodic payments at the point in time one period before the first payment. What is the amount that must be paid (Present Value) for an annuity with a periodic payment

    posted on November 29, 2011
  48. Math

    The formula for calculating a monthly loan payment is R = Pi/[1 - 1/(1+i)^n] where R = the periodic payment, P = the principal, or debt to be paid off, n = the number of payment periods over which the payments will take place, and i = the periodic interest

    posted on November 29, 2011
  49. Algebra

    Letting x, y, and z equal the numbers of .02, .08 and .14 cent stamps. 1-x = 2y + 5 2-z = 3y 3-.02x + .o8y + .14z = 2.26 4-Times 100 yields 2x + 8y + 14z = 226 5-Substituting (1) and (2) into (4) yields ..2(y + 5) + 8y + 14(3y) 6-Can you take it from here?

    posted on November 23, 2011
  50. Algebra II

    The Law of Universal Gravitation states that each particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

    posted on November 22, 2011
  51. gravity

    Addendum to my earlier response. A reduction in velocity of only 1.5m/sec. will drop the space shuttle into an ever so slightly elliptical orbit with an apogee of 6828km, a perigee of 6825.6km. and a period of 5611.86 sec. or 93.53min., allowing the

    posted on November 21, 2011
  52. gravity

    Assume, for discussion purposes, a satellite and the shuttle are in the same 91 minute circular orbit, the satellite 2 minutes ahead of the space shuttle. The shuttle fires its on-orbit thrusters to reduce its orbital velocity. In doing so, the space

    posted on November 20, 2011
  53. physics

    The equations that apply to rising bodies are .....Vf = Vo - gt (the term “g” for acceleration due to gravity is assumed constant on, or near, the surface of the Earth) .....d = Vo(t) - g(t^2)/2 .................2 .....Vf^2 = Vo^2 - 2gd From Vf^2 =

    posted on November 19, 2011
  54. Physics

    L = the left and and The summation of moments about the left end support is 1.4(55) + 3.8(70) - 5R. Solve for R, the right end support load. The summation of vertical forces is L - 55 - 70 + R. Solve for L.

    posted on November 17, 2011
  55. math

    There is an old rule of thumb that states that the time required for an investment to double is equal to 72 divided by the interest rate in percent. Example, for $10,000 to double to $20,000 at 6% interest, compounded annualy, it would have to remain on

    posted on November 17, 2011
  56. math

    Ignoring the corner posts, there are 14 posts along each long side, 9 posts along each short side and the 4 corner posts for a total of 2(14) + 2(9) + 4 = 50

    posted on November 16, 2011
  57. math 140

    This is an ordinary annuity where R dollars is deposited in a bank at the end of each month and earning interest compounded monthly. S(n) = R[(1+i)^n - 1]/i where R = the monthly deposit, S(n) = the ultimate accumulation, n = the number of periods the

    posted on November 16, 2011
  58. Physics

    Since there is no atmosphere in space, the typical control surfaces such as the aileron, rudder, or elevator, used on an airplane would serve no purpose. All spacecraft use what is called an Attitude Control System(ACS) or a Reaction Control System(RCS) to

    posted on November 16, 2011
  59. algebra

    From P R[1 - (1+i)^(-n)]/i where P = present value R = the periodic payment i = the decimal interest per compounding period = I/100 assuming interest compounded anually and n = the number of interest bearing periods. Therefore, P = 7300[1 -

    posted on November 15, 2011
  60. physics

    A rocket is fired in deep space, where gravity is negligible. In the first second it ejects 1/160 of its mass as exhaust gas and has an acceleration of 15.1 m/s^2.What is the speed v_gas of the exhaust gas relative to the rocket? By means of the rocket

    posted on November 15, 2011
  61. math

    Having spent $25, LEa realizes that she still has 60% of her original sum. Therefore, $25 must represent 40% or her original sum making the original sum 25/.40 = $62.50. $62.50 - $25.00 = $37.50. $37.50/$62.50 = .6 or 60%. $3.00/2.6 = $1.15 per liter

    posted on November 15, 2011
  62. Astronomy

    The eccentricity of Mar's orbit is .0934. The semi-major axis is 141,643,675 miles. Therefore, the perihelion distance is r(p) = 141,643,675(1-.0934) = 128,414,418. and r(a) = 141,643,675(1+.0934) = 154,871,931. The mean radius of the earth's orbit is r =

    posted on November 14, 2011
  63. Physics

    From Vf^2 = Vo^2 + 2gh Vf^2 = 0 + 2(9.8)250,000 making Vf = 2213m/s or From h = Vo(t) = g(t^2)/2 250,000 = 9.8t^2/2 making t = 225.87sec. and Vf = Vo + gt Vf = 0 + 9.8(225.87) = 2213m/s

    posted on November 14, 2011
  64. Physics

    Escape velocity derives from Ve = sqrt[2µ/r] where Ve = escape velocity in ft/sec., µ = the gravitational constant of the earth and r = the surface radius. µ can also be stated as GM where G = the gravitational and M = the mass of the planet. Therefore,

    posted on November 13, 2011
  65. PHYSIC

    Since 8000 - 1630 = 6370, the radius of the earth in km, I believe I am on safe ground assuming that you mean "740km" above the earth in your first question. The velocity required to maintain a circular orbit derives from Vc = sqrt(µ/r) where Vc = the

    posted on November 11, 2011
  66. Physics

    Vp/Va = r(a)/r(p)

    posted on November 11, 2011
  67. Math

    How about 43^5.

    posted on November 11, 2011
  68. Math, Finance

    You have the right formula typically shown in the form or Ri = Sn/[(1+i)^n - 1] Sn = 120,000 n = 15x2 = 30 i = .068/2 = .034 Make the keys dance.

    posted on November 10, 2011
  69. Math

    Compound Interest With compound interest, the interest due and paid at the end of the interest compounding period is added to the initial starting principal to form a new principal, and this new principal becomes the amount on which the interest for the

    posted on November 10, 2011
  70. Math

    Might you have noticed the new HD screen ratio comparison to its predecessor? The old screen ratio was 4:3 while the new one is 16:9. That's 4^2/3^2. The new screen ratio actually results in an 11% reduction in actual screen area.

    posted on November 10, 2011
  71. math

    P = 9n n = 1 -2 -3 -4 -5 -6 -7- 8- 9 -10 P = 9-18-27-36-45-54-63-72-81-100 Notice anything?

    posted on November 10, 2011
  72. Math

    If she had 6 left after giving 3/5 to her children, she had 6/(2/5) = 15 to start. If she had 15 left after giving 1/4 to her neighbors, she had 15/(3/4) = 20 to start. If she had 20 left after giving 1/2 to her sister, she had 20/(1/2) = 40 to start out

    posted on November 5, 2011
  73. Physics

    The Law of Universal Gravitation states that each particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

    posted on November 4, 2011
  74. geometry

    Considering all rectangles with a given perimeter, which one encloses the largest area? The traditional calculus approach would be as follows. Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P -

    posted on November 4, 2011
  75. advanced math

    Alternatively: Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area? Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the

    posted on November 4, 2011
  76. advanced math

    Let the sides parallel to the given boundary be x and the sides perpendicular to the boundary be y. Then, 2x + 3y = 168 and A = 2xy. With y = (168 - 2x)/3 A = (336x - 4x^2)/3 dA/dx = 112 - 8x/3 = 0 336 = 8x making x = 84 and y = 28 The area A = 2352m^2.

    posted on November 4, 2011
  77. physics180

    A satellite is in circular orbit at a height R above earth's surface. a)find orbital period. b)what height is required for a circular orbit with a period double that found in part (a)? Allow me to modify your terms. a) The orbit radius R = Re + h where Re

    posted on November 3, 2011
  78. algebra

    There are 132 more silver dollars than 50-cent pieces or D = H + 132. The ratio of silver dollars to 50-cent pieces is 5:2 or D/H = 5/2. How many of each type of coins are in Mr. Hundley's collection? D = 5H/2 Therefore, 5H/2 = H + 132 5H = 2H + 264 3H =

    posted on November 3, 2011
  79. physics

    The average speed for the entire trip is the total distance divided by the total time. Therefore, t1 = 5/6.2 = .806 hr. t2 = 5/V Total distance traveled is 5 + 5 = 10 miles. Total time is t1 + t2 = .806 + 5/V Average driving speed is therefore Vavg =

    posted on November 1, 2011
  80. physics

    The orbital period of a satellite derives from T = 2(Pi)sqrt(r^3/µ) where T = the period in seconds Pi = 3.14 r = the orbital radius in feet and µ = the earth's gravitational constant = 1.407974x10^16 ft^3/sec^2. 139(60)= 2(3.14)sqrt(r^3/1.407974x10^16)

    posted on November 1, 2011
  81. Algebra 2

    Using V = D/T 60D/45 = T - 1 60D/40 = T + 1 Solving for T and equating yields 60(40D - 45D)/1800 = 2 or 5D = 60 making D = 12 miles In order to arrive in 17 minutes, 60(12)/V = 17 yielding = 42.35mph.

    posted on November 1, 2011
  82. Physics

    Given the distance and angle of elevation, the velocity with which the ball left the club face derives from d = V^2(sin(µ))/g which then allows the maximum height reached to be found from h = [V^2(sin^2(µ))]/2g

    posted on October 24, 2011
  83. physics

    How do you determine the altitude at which a satellite must fly in order to complete one orbit in the same time period that it takes the earth to make one complete rotation? The force exerted by the earth on the satellite derives from

    posted on October 24, 2011
  84. Physics

    The period of an orbiting body derives from ....T = 2(Pi)sqrt(r^3/µ) in seconds where T = the period in sec, Pi = 3.1416, r = the orbital radius about the Sun in meters and µ = the planet's gravitational constant = Gm where G = the gravitational constant

    posted on October 23, 2011
  85. math

    For a given perimeter, the rectangle that encloses the maximum area using one side as a building, river, fence, etc. or an adjacent section of fence, is in the ratio of 2/1. Assuming each of the 16 grazing areas is in the length to width ratio of 2/1: Let

    posted on October 20, 2011
  86. PHYSÝCS

    I must admit I am not clear on what you mean by "mean radius" of the earth; average, arithmetic, geometric, harmonic, etc. I am making my own interpretation to be the radius that divides the spherical earth into two equal volumes, ignoring the equatorial

    posted on October 19, 2011
  87. math

    Vf = Vo(t) - 32(t) Vf = 88 - 32t or t = 2.75sec. Height reached from 25 ft. height h = 88(t) - 32(t^2)/2 h = 88(2.75) - 16(2.75)^2 = 121 ft. It takes the same 2.75 sec. to return to the original launch height of 25 feet and an increment more to fall the 25

    posted on October 16, 2011
  88. algebra

    Let the corner pieces be x by x. Then, x(16 - 2x)(9 - 2x) = 120 144x - 50x^2 + 4x^3 = 120 4x^3 - 50x^3 + 144x - 120 = 0 2x^3 - 25x^2 + 72x - 60 = 0 First derivative = 6x^2 - 50x + 72 = 0 3x^2 - 25x + 36 = 0 x = [25+/-sqrt(25^2 - 4(3)36)]/6 x = 1.851

    posted on October 16, 2011
  89. physics

    The weight is 2.0x106 x 2.205 = 4,410,000 lbs. (actually 4,458,000 with a 25,000 lb. payload) Liftoff thrust is 35x10^6N .102 x 2.205 = 7,871,850 lbs. (actually 7,245,000 lbs.) Net liftoff acceleration a = 7,871,850(32.2)/4,410,000 - 32.2 = 25.27fps^2.

    posted on October 15, 2011
  90. Physics 1 - Mechanics

    Being in a 261 mile high orbit, your orbital velocity is 25,125 fps. The orbital period is 92.95 minutes. Assuming that the 21km (68,901 ft.) distance is along the circumference of the orbit, your friend is only 2.75 sec. ahead of you. Clearly, you need

    posted on October 14, 2011
  91. Math WORD PROBLEM

    The actual velocity of a satellite orbiting at 860 miles altitude is 24,113 fps. The period derives from T = 2(Pi)sqrt(r^3/µ) = 113.3 minutes. The actual velocity of a satellite orbiting at 1700 miles altitude is 21,705 fps. The period derives from T =

    posted on October 13, 2011
  92. Algebra

    1--.05x-.6(x+2)=.08 2--Multiplying by 100 yields ...5x - 60x - 120 = 8 3--Collecting terms, -55x = 128 4--Therefore x = -128/55 5--Checking: ...05(-128/55) - .6(-128/55 - 1.2 = .08 ...Multiplying through by 55 yields ....05(-128) - .6(-128) - 66 = 4.4

    posted on October 13, 2011
  93. trig

    There is insufficient information to determine the height after traveling 6000 ft. horizontally. It will definitely be on an angle less than 18º from the launch point.

    posted on October 12, 2011
  94. physics

    Excuse the change in units. 60 year old habits are hard to break. You left us without a clue as to the weight of the rocket. Propellant fractions of rockets typically range from .85 to .90, i.e., the weight of propellant divided by the fully loaded rocket

    posted on October 12, 2011
  95. Physics

    d = (24.5)^2(sin(2*35º)/9.8 = h = (24.5)^2(sin^2(35º)/2(9.8) =

    posted on October 11, 2011
  96. MATH

    Triangular Numbers The number of dots, circles, spheres, etc., that can be arranged in an equilateral or right triangular pattern is called a triangular number. The 10 bowling pins form a triangular number as do the 15 balls racked up on a pool table. Upon

    posted on October 10, 2011
  97. Physics

    g = .379(9.8) = 3.71m/s^2 Up time = down time = 3.1sec. each. Vf = 0 = Vo - 3.71(3.1) making Vo = 11.5m/s h = 11.5(3.1) - 3.71(3.1)^2/2 = 17.82m (a) 17.82m (b) Vo = 11.5m/s.

    posted on October 10, 2011
  98. Algebra

    1--70t1 + 20t2 = 260 2--t1 + t2 = 9 or t1 = 9 - t2 3--Substitute (2) into (1) and solve for t2 4--Distance traveled at 20mph = 20t2

    posted on October 10, 2011
  99. algebra

    The Golden Ratio = 1.618. Therefore, the length would be 1.618(14.72) = 23.817.

    posted on October 10, 2011
  100. algebra

    (30t)^2 + (40t)^2 = 400^2

    posted on October 10, 2011